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3 years ago

How much work is done to increase the speed of a 1.0 kg toy car by 5.0 m/s?

Physics

1 answer:

soldi70 [24.7K]3 years ago

6 0

Answer:

The correct option is (b).

Explanation:

We need to find the work done to increase the speed of a 1 kg toy car by 5 m/s.

We know that, the work done is equal to the kinetic energy of an object i.e.

$W=\Delta K\\\\W=\dfrac{1}{2}mv^2\\\\W=\dfrac{1}{2}\times 1\times 5^2\\W=12.5\ J$

So, 12.5 J of work is done to increase the speed of a 1.0 kg toy car by 5.0 m/s.

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Which measure of central tendency refers to the average of all the scores in a data set?

dedylja [7]

Answer

Mean

Explanation

For any set of date there is a typical value for the probability distribution called central tendency.

There three types of central tendency mean, median and mode.

Mean is the average value of all data set in a distribution.

median is the the central value when the data is arranged in ascending or descending order.

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3 years ago

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A flare is launched from a life raft with an initial velocity of 192 ft/sec. How many seconds will it take for the flare to retu

DIA [1.3K]

We use the formula,

$h= ut- 16 t^2$

Here, h is the variable represents the height of the flare in feet when it returns to the sea so, h = 0 and u is the initial velocity of the flare, in feet per second and its value of 192 ft/sec.

Substituting these values in above equation, we get

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3 years ago

Which phenomenon occurs when the moon and Earth are aligned with the sun? A. Retrograde motion B. Solstice C. Equinox D. Eclipse

muminat

Answer: Eclipse

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3 years ago

A hot air balloon is traveling vertically upward at a constant speed of 4.5 m/s. When it is 28 m above the ground, a package is

ella [17]

Answer:

1.97 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow 21=4.5t+\frac{1}{2}\times -9.8\times t^2\\\Rightarrow 21=-4.5t-4.9t^2\\\Rightarrow 4.9t^2+4.5t-28=0\\\Rightarrow 49t^2+45t-280=0$

Solving the above equation we get

$t=\frac{-45+\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}, \frac{-45-\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}\\\Rightarrow t=1.97, -2.89$

So, the time the package was in the air is 1.97 seconds

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A roller coaster has a mass of 275 kg. It sits at the top of a hill with height 85 m. If it drops from this hill, how fast is it

Minchanka [31]

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2 years ago

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